toán 11- giải bất phương trình

Question

Giải BPT sau

$4\sqrt{x+1} +2\sqrt{2x+3} \leq (x-1)(x^{2} -2)$

Confusion · Answer 1 · 6/16/2016 3:06:03 PM

Condition:

$x\geq -1$

Disequations

$\Leftrightarrow 4(\sqrt{x+1}-2)+2(\sqrt{2x-3})\leq x^3-x^2-2x-12$

$\Leftrightarrow \frac{4(x-3)}{2+\sqrt{x+1}}+\frac{4(x-3)}{3+\sqrt{2x+3}}-(x-3)(x^2+2x+4)\leq 0$

$(x-3)$

$[\frac{4}{2+\sqrt{x+1}}+\frac{4}{3+\sqrt{2x+3}}-(x+1)^2-3]$

$\leq 0$

$+)x=-1$ is satisfy.

$+)x>-1\rightarrow$ Blue

$<\frac{4}{2+0}+\frac{4}{3+1}-0-3=0$

$\rightarrow ...............$

$\rightarrow x\geq 3.$

Combined with condition, we get:

$x=-1$ v

$x\geq 3$

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